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March 1,1868. J THE CIVIL ENGINEER AND ARCHITECT’S JOURNAL. 83 chalk the induration will be enabled to proceed rapidly by the abstrac tion of carbonic acid by the hydrate of lime from the chalk. Of course, when the chemists deny this, we are quite at fault; but there is still this fact to be accounted for, namely, that when, as I am told, an old limestone wall is pulled down, the mortar joint is found to adhere to the limestone surfaces with a tenacity which is otherwise unexampled; indeed, it was upon the observation of this fact that Mr. Westmacott based his theory. Possibly Professor Ansted may be able to suggest a reason for this. (To be concluded in our next.') THE STABILITY OF DOMES* By E. Wyndham Tarn, M.A., Architect. In my former paper on this subject I obtained formulse for calculating the thrust of a spherical dome of uniform thickness, by supposing it to consist of a number of thin ribs, each of which is formed by two vertical planes intersecting at the axis of the dome, and making a small angle (</>) with each other, and then treating each rib as forming, with the corresponding one on the opposite side, a complete arch. I now propose to apply the same method to domes of other forms than the spherical. The “ Gothic ” dome, of which that of the cathedral at Florence is a splendid example, has for its section a pointed arch formed by two segments of circles, the centre from which each is struck being on the springing line. Fig. 4, Plate 4, represents a rib or wedge cut out of a dome’ of this kind, 0 being the centre from which the arc B E D is struck. I will suppose the the thrust N of the corresponding rib on the opposite side to act at n, so that On=i (R-|-r), R and r being the outer and inner radii of the dome, and the line On making the angle a with the vertical. Let E F be any joint in the masonry, making at O the angle d with On ; P the weight of the portion of the rib above E F ; F that of the portion between E F and A B. Suppose g to be the centre of gravity of the upper part of the rib whose weight is P, and G its projection on the springing line 0 A. As in my former paper, I take the moments of N and P about E, namely, P® and Ny, which must be equal when there is equilibrium, or N = P^. I now proceed to express N in terms of d, and then by calcu lating the value of N for various values of d, I find the point on the arch at which the thrust N is greatest. The surface of the dome (which is circular on plan) is formed by the revolution about the vertical line C Z of the arc BED; and the rib, which forms the portion under consideration, I take as subtending an angle (^>) of 2° at the axis; so that </> ='0.34907 circular measure. Using the same notation as in my former paper, I find (putting 8 as the weight of one cubic foot of masonry) (sin (a+d)—sin a) dr . dd . d<j> .... (A) = (R 3 —r 3 ) (cos a—cos (a+d)—d . sin a) the limits of r being r and R, and those of d being 0 and d. And calling GZ=z, we have Z_ AT* -3 (sin (a+d)—sin a) 2 . cos <f> . dr . dd . df> fffr 1 (sin (a+d)—sin a) dr . dd . df> _sin Jfr 3 (sin (a-|-d)—sin a)°dr . dd JJp (sin (a+d)—sin a) dr . dd '' and <j> being a small angle, we can put sin i<f> _ Integrating between the same limits, we find * Concluded from page 37 ante. z=?S^ X 4 k- r 3 id (1+2 sin S d)'+J (sin 2a—sin 2 (d+d)) —2 sin a (cos a_cos (a+d)) cos a—cos (a+d)—d . sin a Calling x the lever arm of P about E, and y that of N, we have x=r (sin (a+d)—sin a)—Z y = i (R-|-r). cos a—r . cos (a+d) In order to apply these formulae to a dome of any particular form, we must know the value of the angle a. In the Florentine example this angle is 22J°, or a=+=-3927, sin a='38268, cos a='92388, sin 2a='7O711, |(l+2 sin 2 a)='64645. Therefore with this value of a P = 3.0-’ (R 3 —r 3 ) ('92388—cos (a+d)—'38268 d) P . x—8. <f> (1 (R 3 —r 3 ) r ('92388—cos (a+d)—'38268 d) (sin (a-j-d)—-38268)—| (R 4 —r 4 ) (-64645 d+| ('70711—sin 2(a-j-d)) —•76536 ('92388—cos (a+d))) | i/ = .46194 (R-j-r)—r . cos (a+d). In order to find the value of d which makes N a maximum, I put R = ll ft., r=10 ft., and calculate N=P- for several values V of d until the greatest value of N is obtained. Proceeding in this manner I find that N is greatest when d is an angle of 54°, or when the joint E F makes an angle of 13J° with the springing line, since a-\-d = 76|°. Substituting this value for d in the foregoing expressions, I find that N is greatest when P =-003837 3 (R 3 —r 3 ), and _ -002262 (R 3 —r 3 ) r—-001258 (R 4 —r 4 ) “ -46194 R-f--22849 r ---(C) We can in the same way find the value of F from the integral (A), taking limits from d=54° to d=90—a, which gives F=—3^ (B 3 -—r 3 ) cos (a+d)—(*—a—d) sin a j = -5— (R 3 —r 3 ) (-14328) = -001667 3 (R 3 —r 3 ). Now, to find the necessary thickness of pier to resist the thrust of the dome, we have to take moments of all the forces about S, the outer edge of the pier ; and since the moments of N and P balance about E, we can, by the principles of mechanics, remove N and P to act at E in their or ginal directions, and take their moments about S. Calling N . b the moment of N about S, P . a that of P, F . c that of F, Q . q that of the pier itself; we have (as explained in my former paper) the Equation of Stability. 2N6=P . «+Q . </+F . c from which we obtain the value of t, the thickness of the pier. The value of Q. q was found in my former paper to be (putting I for the half span B Z) Q . q=-005818 8! (3st 2 +< 3 ) H, II being the height of the pier, and 3, its weight per cubic foot. a=Se=H-r (1—sin (a+d)) = <+-02763 r &=Si/=H-j-r . cos (a-|-d) = H-J-'23345 r If we call Z, the horizontal distance from the axis of the centre of gravity of the portion of the rib below E F, the value of Z, is found from the integral (B), taking the same limits of d as in finding the value of F; so that 3 pp— r < i (1+2 sin 2 a) (^—a—d)+| sin 2 (a i d)—2 sin a cos (a+d) Z i=4R e t- 3 ———— —————— dos (a+d) — ( 9 —a—d) Sin a And putting for a and d their values as above 3 R 4 —r 4 -08713 Z1 _ 4 R 3 —r ' 1 -14328’ and c=t-\-s—Z x Therefore F . c='OO1667 8 (R 3 —r 3 ) (t+s)—'00076 3 (R 1 —r 4 ) P. a=-003837 8 (R 3 —r 3 ) < + '0001059 8 (R 3 —r 3 ) r. In order to compare the thrust of a dome of this form with that of a spherical dome of equal span and thickness, I will apply these formula; to the example in which the span is 20 ft. and thickness of dome 1 ft, having in my former paper calculated the thrust of a spherical dome of these dimensions. Here we have s=10 ft., r= 1(> +\ sm a =16 -5 ft., R=i7-5 ft., H=50 ft., ’ 1—sin a 8 = 125 lbs., 8, = 150 lbs.
